Triangular Numbers

Date: 07/07/98 at 08:09:35
From: frances
Subject: Triangular numbers
Dear Dr. Maths,
How do you know a number is a triangular number or not, and how is
n/2(n+1) derived - i.e. how did they get it?
Frances

Date: 07/07/98 at 09:41:36
From: Doctor Rob
Subject: Re: Triangular numbers
A number N is a triangular number if 8*N + 1 is a perfect square.
Example:
45 is a triangular number because 8*45 + 1 = 361 = 19^2 is a perfect
square.
88 is not a triangular number because 8*88 + 1 = 705 = 3*5*47 is not a
perfect square.
Draw a triangle of size n. Draw another next to it, upside down.
Together they form a rectangle n by n+1. Here is an example of a size
6 triangle, two of which form a 6-by-7 rectangle:
o oooooo
oo ooooo
ooo oooo
oooo ooo
ooooo oo
oooooo o
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/

Date: 07/08/98 at 03:41:18
From: Frances
Subject: Use of triangular numbers
Dear Dr Maths,
I am presently writing an oral report on triangular numbers. I need to
know the purpose of these numbers - i.e. what are they used for?
Also, where does the 8 in '8*N + 1' when finding whether N is a
triangular number come from?

Date: 07/08/98 at 10:47:51
From: Doctor Rob
Subject: Re: Use of triangular numbers
If a triangular number has the form N = n*(n+1)/2, then 2*N = n^2 + n.
Now I "complete the square" on the righthand side by adding 1/4:
2*N + 1/4 = n^2 + n + 1/4 = (n+1/2)^2.
Now to clear fractions and get everything in terms of whole numbers,
I multiply by 4: 4*(2*N+1/4) = 8*N + 1 = 4*(n+1/2)^2 = (2*n+1)^2.
Another way to look at this is to take the equation N = n*(n+1)/2, and
use the Quadratic Formula to solve it for n:
(1/2)*n^2 + (1/2)*n - N = 0,
n = (-1/2 +- sqrt[1/4 - 4*(1/2)*(-N)])/(2*[1/2]),
= -1/2 +- sqrt[1/4 + 2*N],
= (-1 +- sqrt[8*N + 1])/2.
In order for n to be a rational number, 8*N + 1 must be a perfect
square, and if it is a perfect square, it is the square of an odd
whole number, so n is a whole number, too.
One of the uses of triangular numbers is to count things. How many
ways can you pick two objects out of a set of n objects, if order
doesn't matter? The answer is the (n-1)st triangular number.
Example:
From {a,b,c,d,e} we can pick ab, ac, ad, ae, bc, bd, be, cd, ce, or
de. There are 4 of these starting with a, 3 starting with b,
2 starting with c, and 1 starting with d. The total number is
10 = 4 + 3 + 2 + 1. There are 10 ways of choosing 2 objects from a set
of 5, and 10 = 4*5/2, the 4th triangular number.
By the way, this is how you count the number of edges and diagonals of
a pentagon. There are 10 of them, 5 edges and 5 diagonals (pick two of
the vertex points in all 10 ways possible, and connect them with line
segments).
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/